In 1851, a French physicist named Jean-Bernard-Leon Foucault suspended an iron ball with a radius of approximately 0.5 feet from the ceiling of the Pantheon in Paris with a wire that was over 200 feet long. The ball was used as a pendulum, and it could swing more than 12 feet back and forth. Beneath the ball he placed a circular ring with sand on top of it. Attached to the bottom of the ball was a pin, which scraped away the sand in its path each time the ball went by. To get the ball started on a perfect plane, the ball was held to the side by a cord until it was motionless. At that point, the cord was burned, which started the ball swinging. As the ball continued to swing as a pendulum, the path the pin carved into the sand changed, as the floor itself, as well as the rest of the Earth, was moving beneath it.

Essentially, the Foucault pendulum demonstrates the rotation of the Earth. The Foucault pendulum is not forced to stay in a fixed plane like Newton’s pendulum, also known as Newton’s cradle, which means it can move freely in response to the Coriolis force. The Coriolis force, also known as the Coriolis effect, occurs when masses above the Earth’s surface, such as a bullet or rocket, appear to be deflected from their trajectory, meaning they don’t reach their intended location straight ahead of them. In fact it is our frame of reference, the Earth, which is changing. Our frame of reference changes due to our uniform circular motion around the Earth. As the Earth is not a perfect circle (elliptical), the closer to the equator you are, the further away you are from the Earth’s centre and the less force of gravity you experience. The Earth’s radius is 6378 km. As a result of your increased distance from the centre of the Earth at the maximum point at the Earth’s radius on the equator, you have a lower centripetal force at that location. This is shown by the formula for centripetal acceleration, which is:

...UniformCircularMotion
PES 115 Report
Objective
The purpose of this experiment is to determine the relationships between radiuses, mass, velocity and centripetal force of a spinning body. We used logger pro to accurately measure the orbital period of the spinning mass and used these measurements to determine the interrelated interactions of the specified properties and viewed the results graphically.
Data and Calculations
The black markings on the string are about 10 cm apart in length, measured from the center of the spinning mass.
Part A: Factors that influence CircularMotion
Velocity versuse Centripetal Force
Fill out the table holding the Spinning mass (M) and the radius (R) constant.
Figure 1: Experimental setup for the lab
Which Spinning Mass did you select _hook with foam wrapping_ (Tennis ball, etc..)
What is the mass of the Spinning mass _0.0283_ kg.
What Radius did you select _0.30_ m (around 20 cm is a good choice).
Fill out the tables for five different hanging mass values.
Hanging Mass (m) [kg]
0.1001 kg
0.1992 kg
0.2992 kg
0.4000 kg
0.4997 kg
Revolution Number and Time per Revolution (T) [sec]
1
0.61337 s
0.413210 s
0.367288 s
0.316510 s
0.271455 s
2
0.613087 s
0.403737 s
0.370600 s
0.310189 s
0.274200 s
3
0.613727 s
0.393689 s
0.374100 s
0.316308 s
0.273700 s
4
0.611319 s
0.39364 s
0.368047 s
0.309619 s
0.279400 s
5
0.618954 s
0.388600 s
0.365853 s
0.300742 s
0.282000 s...

...
E105: UNIFORMCIRCULARMOTION
NADONG, Renzo Norien D.
OBJECTIVE
The purpose of this experiment is to quantify the centripetal force on the body when one of the parameters is held constant and to verify the effects of the varying factors involved in circularmotion. Mainly, horizontal circular type of motion is considered in this activity.
Circularmotion is defined as the movement of an object along the circumference of the circle or the manner of rotating along a circular path. With uniformcircularmotion it is assured that the object traversing a given path maintains a constant speed at all times. Centripetal force is a force that tends to deflect an object moving in a straight path and compels it to move in a circular path.
MATERIALS AND METHODS
This experiment was divided into three parts in order to further study and observe the factors that affect the centripetal force of a body. The concept of this experiment is the same on all parts, which is getting the centripetal force given with three different conditions. Every part of the experiment was executed just the same. Mass hanger plus a desired mass of weights were hanged over the clamp on pulley to determine a constant centripetal force which will act as the actual value. But on the third part of this...

...Exploration Guide: UniformCircularMotion
Go to www.explorelearning.com and login. Please type or write your answers on a separate sheet of paper, not squished in the spaces on these pages. When relevant, data collected should be presented in a table.
Objective: To explore the acceleration and force of an object that travels a circular path at constant speed. Motion of this kind is called uniformcircularmotion.
Part 1: Centripetal Acceleration
1. The Gizmotm shows both a top view and a side view of a puck constrained by a string, traveling a circular path on an air table. Be sure the Gizmo has these settings: radius 8 m, mass 5 kg, and velocity 8 m/s. Then click Play and observe the motion of the puck.
a. The puck in the Gizmo is traveling at a constant speed, but it is NOT traveling at a constant velocity. Explain why.
b. Because the velocity of the puck is changing (because its direction is changing), the puck must be experiencing an acceleration. Click BAR CHART and choose Acceleration from the dropdown menu. Check Show numerical values. The leftmost bar shows the magnitude of the acceleration, or |a|. (The other two bars show the x- and y-components of the acceleration, ax and ay.) What is the value of |a|? Jot this value down, along with radius = 8 m, so that you can refer to it later.
c. Keeping velocity set...

...UniformCircularMotion – a constant motion along a circle; the unfirom motion of a body along a circle
Frequency (f) – the number of cycles or revolutions completed by the same object in a given time; may be expressed as per second, per minute, per hour, per year, etc.; standard unit is revolutions per second (rev/s)
Period (T) – the time it takes for an object to make one complete revolution; may be expressed in seconds, minutes, hours, years, etc.; standard unit is seconds per revolution (s/rev)
Note: Period and frequency are reciprocals: T = 1/f; f = 1/T.
Sample Problems:
1. Suppose the rear wheel makes 5 revolutions in 1 minute. Find the wheel’s period and frequency.
2. As a bucket of water is tied to a string and spun in a circle, it made 85 revolutions in a minute. Find its period and frequency.
3. * An object orbits in a circularmotion 12.51 times in 10.41 seconds. What is the frequency of this motion?
Tangential Speed (v or vs) – average speed; rotational speed; speed of any particle in uniformcircularmotion; standard unit is meters per second (m/s); v = Cf = C/T = 2πrf = 2πr/T = rω
Sample Problems:
3. What is the rotational speed of a person standing at the earth’s equator given that its radius is 6.38*106 m and that it takes 365 days for the earth to complete a revolution?
4. A ball...

...changing. Because the direction is changing, there is a ∆v and ∆v = vf
- vi
, and
since velocity is changing, circularmotion must also be accelerated motion.
vi
∆v vf
-vi
vf2
If the ∆t in-between initial velocity and final velocity is small, the direction of ∆v
is nearly radial (i.e. directed along the radius). As ∆t approaches 0, ∆v becomes
exactly radial, or centripetal.
∆v = vf
- vi
vi
vf
vf
∆v
-vi
Note that as ∆v becomes more centripetal,
it also becomes more perpendicular with vf
.
Also note that the acceleration of an object depends on its change in velocity ∆v;
i.e., if ∆v is centripetal, so is ‘a’.
From this, we can conclude the following for any object travelling in a circle at
constant speed:
The velocity of the object is tangent to its circular path.
The acceleration of the object is centripetal to its circular path. This type of
acceleration is called centripetal acceleration, or ac
.
The centripetal acceleration of the object is always perpendicular to its
velocity at any point along its circular path.
v
ac
ac
v 3
To calculate the magnitude of the tangential velocity (i.e., the speed) of an
object travelling in a circle:
• Start with d = vavt where ‘vav’ is a constant speed ‘v’
• In a circle, distance = circumference, so d = 2πr
• The time ‘t’ taken to travel once around the circular...

...Title: UniformCircularMotion
Objective: To investigate the relationship between FnetT² and radius
Proposed Hypothesis: FnetT² is directly proportional to the radius
Manipulated variable: Radius of the circularmotion
Responding variable: The time taken for 20 rotations
Controlled variables: The mass of the rubber stopper, the mass of the weight hanger, the total weight of the slotted weight, the length of the PVC tube
Apparatus and Materials: rubber stopper, stopwatch, weight hanger, slotted weights,
crocodile clip, metre rule, thread, PVC tube
Diagram:
Procedure: 1. Weigh and record the masses of weight hanger and rubber stopper.
2. Tie the thread to the rubber stopper.
3. Pass the thread through the PVC tube.
4. Tie a node at the end of the thread and hang the weight hanger which is with
0.08g slotted weights on it.
5. Measure the 0.1m radius from the bottom of PVC tube and mark it.
6. Zheng Yie starts to rotate the thread with an acceleration until the bottom of
PVC tube is reached the mark.
7. Keep the speed of rotation constant so that the bottom of PVC tube is
always touched the mark.
8. After the speed is kept at constant speed, Adeline starts to studies the time
taken for 20 rotations by using a stopwatch.
9. Adeline is also responsible to record the time taken for 20 rotations.
10. Step 5...

...INVESTIGATING CIRCULARMOTION 11/3/04
AIM
To examine some of the factors affecting the motion of an object undergoing uniformcircularmotion, and then to determine the quantitative relationship between the variables of force, velocity and radius.
APPARATUS
Rubber bung Metre rule 50 gram slot masses
Glass tube 50-gram mass carrier 50-gram slot masses Metre rule
Stopwatch Sticky tape Metre rule String
THEORY
As in Jacaranda HSC Science Physics 2 p.54
In this experiment when the rubber bung is moving in a circularmotion and the string it is tied to moves neither up or down a constant radius is being maintained. For this to be true the centripetal force must equal the gravitational force hence
Mv"/r = mg from this
v"/r =mg/M and v" ∞ r therefore as v increases so does r and vice versa.
Where
m = Mass of mass carrier + masses (kg)
g = acceleration due to gravity 9.8 m/sec"
M = mass of object in motion (kg)
v = instantaneous velocity of mass (m/sec)
r = radius of circularmotion (m)
METHOD
As in Jacaranda HSC Science Physics 2 p.54
However instead of measuring the time for 10 revolutions, the time for 20 revolutions was measured, this allowed more accurate results to be obtained. Furthermore the lengths given in the book were used as merely guidelines and not followed precisely also 50 and 100-gram masses...

...Term 3
UniformCircularMotion
When a body moves in a circular path with a constant speed, it is said to undergo uniformcircularmotion.
Although the speed is constant, velocity is continually changing, since it is constantly changing its direction of motion.
Centripetal
V
V
ac
ac
Acceleration is directed towards the centre of the circle and is therefore called “centripetal acceleration.”
ac =v^2r
ac =v^2r
If T is the time taken for one revolution then: V = 2πrT
ac =v^2r
= (2πrT)^2r
= 4π^2r^2T^2r
= 4π^2rT^2
Let us work with ac against T^2 on graphs.
Note:
Frequency is the number of revolutions completed per seconds. Cycles/seconds
Centripetal Force
… is the force responsible for centripetal acceleration. Hence the centripetal force may be defined as the force directed towards the centre of the circular path responsible for restraining the object to undergo uniformcircularmotion.
Fc= m.v^2r
Example
A car of mass 1.20X10^3 kg is rounding a curve of radius 1.55X10^3m at a speed of 25.0m/s/
Find: 1. The centripetal acceleration.
2. the centripetal force
ac =v^2r
Non-UniformCircularMotion
Vertical Circle
Resultant force toward the centre is
FNET = FC= mv2r
Top: mv2r = mg +...